# Energy in harmonic motion

1. Nov 17, 2007

### gills

1. The problem statement, all variables and given/known data
Problem 60 in attached image:

2. Relevant equations

E = U + K
E = $$\frac{1}{2}$$kx$$^{2}$$ + $$\frac{1}{2}$$mv$$^{2}$$ + $$\frac{1}{2}$$Iw$$^{2}$$

w = $$\sqrt{\frac{k}{m}}$$

x(t) = Acos(wt)

Equation 13.3 they are referring to in the book:

m($$\frac{d^{2}x}{dt^{2}}$$) = -kx ----> which = --->

$$\frac{d^{2}x}{dt^{2}}$$ + $$\frac{k}{m}$$x = 0

This is what i'm working with now...am i on the right track?

2. Nov 17, 2007

### gills

Ok if i look at problem 59 as a helper up top and differentiate that equation with respect to time, i end up with:

E = $$\frac{1}{2}$$mv$$^{2}$$ + $$\frac{1}{2}$$kx$$^{2}$$ ---> E is a constant so that goes to 0.

v = $$\frac{dx}{dt}$$ which = $$\frac{d^{2}x}{dt^{2}}$$ when differentiated

but if you differentiate x, that is $$\frac{dx}{dt}$$. I'm confused as to how they end up with solely x in that equation. So by what I solved it'll look like this:

0 = m$$\frac{d^{2}x}{dt^{2}}$$ + k$$\frac{dx}{dt}$$

if i rearrange that, i more or less get the eq. 13.3, but it's not exactly. What is being done wrong?

m$$\frac{d^{2}x}{dt^{2}}$$ = - k$$\frac{dx}{dt}$$

3. Nov 17, 2007

### gills

Ok, i think i've got it. I didn't use the chain rule with differentiating with respect to t:

E = $$\frac{1}{2}$$m$$\frac{dx}{dt}$$$$^{2}$$ + $$\frac{1}{2}$$kx$$^{2}$$ --->

0 = m$$\frac{dx}{dt}$$ * $$\frac{d^{2}x}{dt^{2}}$$ + kx * $$\frac{dx}{dt}$$

m$$\frac{dx}{dt}$$ * $$\frac{d^{2}x}{dt^{2}}$$ = - kx * $$\frac{dx}{dt}$$ ---> $$\frac{dx}{dt}$$ cancels on both sides and we end up with --->

m$$\frac{d^{2}x}{dt^{2}}$$ = - kx ---> which is the eq they asked for

Now I'll move onto 60 if that's correct. Anyone have insight?

4. Nov 17, 2007

### gills

Ok, based on what i solved up ^^ there:

using eq that i posted in the original post:

E = $$\frac{1}{2}$$kx$$^{2}$$ + $$\frac{1}{2}$$mv$$^{2}$$ + $$\frac{1}{2}$$Iw$$^{2}$$ --->

w = $$\frac{v}{r}$$ in rotational motion.

I of a solid disk = $$\frac{1}{2}$$mr$$^{2}$$ --->

$$\frac{1}{2}$$Iw$$^{2}$$ = $$\frac{1}{2}$$($$\frac{1}{2}$$mr$$^{2}$$ * ($$\frac{v}{r}$$)$$^{2}$$ --->

r's cancel each other out we end up with $$\frac{1}{4}$$mv$$^{2}$$ --->

E = $$\frac{1}{4}$$m$$\frac{dx}{dt}$$$$^{2}$$ + $$\frac{1}{2}$$m$$\frac{dx}{dt}$$$$^{2}$$ + $$\frac{1}{2}$$kx$$^{2}$$ --->

E = $$\frac{3}{4}$$m$$\frac{dx}{dt}$$$$^{2}$$ + $$\frac{1}{2}$$kx$$^{2}$$ ---> now differentiate w.r.t t on both sides--->

$$\frac{3}{2}$$m$$\frac{dx}{dt}$$ * $$\frac{d^{2}x}{dt^{2}}$$ = -kx * $$\frac{dx}{dt}$$ ---> $$\frac{dx}{dt}$$ cancels on both sides and we end up with --->

$$\frac{3}{2}$$m$$\frac{d^{2}x}{dt^{2}}$$ = -kx

so the only difference between energy in a regular oscillator and one connected to a wheel that's not sliding is $$\frac{3}{2}$$?? Something doesn't seem right. Anyone??

5. Nov 18, 2007

### gills

bump^^^

6. Nov 18, 2007

### Staff: Mentor

Looks good to me!

The spring hasn't changed, but now it has to provide the energy to roll the cylinder as well as slide it. So the same mass "feels" heavier.

It may be instructive to analyze this problem directly in terms of the forces involved and verify that you get the same answer. Realize that static friction acts to oppose slipping between the cylinder and the surface.

7. Nov 18, 2007

### gills

ok, i'll look at it that way.

moving onto the second part of the problem where it asks for angular frequency (w):

k = mw$$^{2}$$ ---> sub into derived formula the solve for w --->

$$\frac{3}{2}$$m$$\left(\frac{d^{2}x}{dt^{2}}\right)$$ = -(mw$$^{2}$$)x ---> m's cancel on both sides

w = - $$\sqrt{\left(\frac{3\frac{d^{2}x}{dt^{2}}}{2x}}\right)$$

how does that look? w has a negative sense?

8. Nov 18, 2007

### Staff: Mentor

The way to get angular frequency is by comparing your new differential equation with the standard differential equation, which is:
$$d^{2}x/{dt^{2} = -(k/m)x$$

From the solution to that equation:
$$\omega = \sqrt{k/m}$$

Realizing that k and m are just constants, what would be the angular frequency for the cylinder problem?

(Useful rule of thumb to drum into your brain: The same equations have the same solutions. )

9. Nov 18, 2007

### gills

Doc,

I know it should be easy to see, but i can't see it right now.

Is it just going to be w = 3/2 * $$\sqrt{k/m}$$ ?? give me a few more pointers...

10. Nov 18, 2007

### gills

am i using the equation x(t) = Acos(wt) and pluging that into x and differenting it twice to get $$d^{2}x/{dt^{2}$$, and then pluggin that in?

Maybe i'm having trouble understand your post because i'm new to differential equations. Any guidance would be great. Thanks.

11. Nov 18, 2007

### Staff: Mentor

Almost, but not quite right.

If the solution to the standard equation:

$$d^{2}x/{dt^{2} = -(k/m)x$$

Has:

$$\omega = \sqrt{k/m}$$

Then what's the solution to this equation:

$$d^{2}x/{dt^{2} = -(2k/3m)x$$

Remember that k/m is just a constant as far as the equation is concerned. What's the new constant?

12. Nov 18, 2007

### gills

w = $$\sqrt{(2k/3m)}$$ ??

If that's the case, can you please explain or show me how that's the solution of the original equation. Thanks a lot for you help!

13. Nov 18, 2007

### Staff: Mentor

Right!

That's just the value for $\omega$, not the complete solution. (Sorry if I wasn't clear.) The solution is what you quoted before:
x(t) = Acos(wt) [plus a constant or phase factor, of course]

Plug that into the equation and verify that it's a solution.

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